Answer
a.) There are five $sp^{2}$ hybridized carbon atoms.
b.) The hybridization of the oxygen atom is $sp^{2}$.
c.) The C-O double bond is formed from $C_{sp^{2}}-O_{sp^{2}}$.
d.) Cucumber aldehyde contains 23 sigma bonds.
e.) Cucumber aldehyde contains 3 pi bonds.
Work Step by Step
a.) Count the number of carbon atoms with a double bond. All carbon atoms with one double will have a hybridization of $sp^{2}$.
b.) To find the hybridization, count the number of groups surrounding oxygen (4 groups = $sp^{3}$; 3 groups = $sp^{2}$; 2 groups = $sp$).
c.) To figure out the orbitals being used, you must first know the hybridization of each atom included in that bond. To find the hybridization, count the number of groups surrounding the carbon and oxygen atom found in the carbon-oxygen double bond.
d.) Count the number of bonds the molecule has. All single, double and triple bonds contain one sigma bond each.
e.) Count the number of double bonds. Only double and triple bonds contain pi bonds. Single bonds do not contain any pi bond.
